Jumping Exponents

Let $S_n = x^n + y^n + z^n$ and $P(a) = (a-x)(a-y)(a-z) = a^3 + Aa^2 + Ba + C = a^3 + Ba + C$ .

Thus, we have $\begin{aligned} x^n + Bx^{n-2} + Cx^{n-3} & = 0 \\ y^n + By^{n-2} + Cy^{n-3} & = 0 \\ z^n + Bz^{n-2} + Cz^{n-3} & = 0 \end{aligned}$ Summing yields $S_n = -BS_{n-2} - CS_{n-3}$ or just by the fact that $a^3 + Ba + C = 0$ is the characteristic equation.

We can calculate that $\begin{aligned} S_2 & = -2B \\ S_3 & = -3C \\ S_4 & = 2B^2 = 18 \\ S_5 & = 5BC = 30 \end{aligned}$ which implies that $B = \pm 3$ . However, note that $S_2 = -2B \geq 0$ which means that $B = -3$ and $C = -2$ .

$x,y,z$ are the solutions to the polynomial $a^3 -3a - 2 = 0$ . Thus, the solutions are $(2, -1, -1)$ and its permutations